9.3 Clausius-Clapeyron equation

Clausius-Clapeyron Equation

The Clausius-Clapeyron equation connects vapor pressure to temperature along a phase boundary. It lets you predict how vapor pressure changes with temperature and calculate enthalpies of vaporization or sublimation from pressure-temperature data. The derivation flows directly from the condition of phase equilibrium and the definition of chemical potential.

Derivation of the Clausius-Clapeyron Equation

The derivation starts from the fact that, at phase equilibrium, the chemical potential must be equal in both phases. Here's how it builds, step by step:

1. Start with the differential of chemical potential for a pure substance: $d\mu = -s\,dT + v\,dP$

2. Set the two phases equal. At equilibrium between liquid and vapor, $\mu_l = \mu_v$, so any infinitesimal change along the coexistence curve must satisfy: $d\mu_l = d\mu_v$

3. Substitute the chemical potential expression for each phase: $-s_l\,dT + v_l\,dP = -s_v\,dT + v_v\,dP$

4. Rearrange to isolate $dP/dT$: $\frac{dP}{dT} = \frac{s_v - s_l}{v_v - v_l} = \frac{\Delta s}{\Delta v}$

   This is the Clapeyron equation in its exact form. It applies to any phase transition with no approximations.

5. Relate the entropy change to latent heat. During a reversible phase transition at constant $T$ and $P$: $\Delta s = s_v - s_l = \frac{L}{T}$

where $L$ is the molar latent heat (enthalpy of vaporization or sublimation).

1. Apply two approximations to get the Clausius-Clapeyron form:

   • The molar volume of vapor is much larger than that of the condensed phase: $v_v \gg v_l$, so $\Delta v \approx v_v$
   • The vapor behaves as an ideal gas: $v_v = RT/P$

2. Substitute into the Clapeyron equation: $\frac{dP}{dT} = \frac{LP}{RT^2}$

3. Separate variables to get the standard differential form: $\frac{dP}{P} = \frac{L}{RT^2}\,dT$

   Or equivalently: $\frac{d(\ln P)}{dT} = \frac{L}{RT^2}$

Notice the distinction: the Clapeyron equation (step 4) is exact, while the Clausius-Clapeyron equation (step 8) relies on the ideal gas and negligible liquid volume approximations.

Vapor Pressure Calculations

To find the vapor pressure at a new temperature, you integrate the Clausius-Clapeyron equation assuming $L$ is approximately constant over the temperature range:

1. Integrate both sides between states 1 and 2: $\int_{P_1}^{P_2} \frac{dP}{P} = \frac{L}{R} \int_{T_1}^{T_2} \frac{dT}{T^2}$

2. Evaluate the integrals: $\ln\frac{P_2}{P_1} = -\frac{L}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$

3. Solve for $P_2$: $P_2 = P_1 \exp\left[-\frac{L}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)\right]$

To use this, you need one known vapor pressure $P_1$ at temperature $T_1$, plus the latent heat $L$. Common reference points include water's normal boiling point ($P = 1\,\text{atm}$ at $T = 373\,\text{K}$, $L_{vap} \approx 40.7\,\text{kJ/mol}$) or ethanol's ($P = 1\,\text{atm}$ at $T = 351.5\,\text{K}$).

Enthalpy Determination from Clausius-Clapeyron

You can run the calculation in reverse: if you have vapor pressure measurements at two temperatures, you can extract the latent heat.

1. Start from the integrated form: $\ln\frac{P_2}{P_1} = -\frac{L}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$

2. Solve for $L$: $L = -R\,\frac{\ln(P_2/P_1)}{1/T_2 - 1/T_1}$

3. Plug in your measured $(T_1, P_1)$ and $(T_2, P_2)$ values.

The result gives you $L_{vap}$ if you're looking at a liquid-vapor boundary, or $L_{sub}$ if you're looking at a solid-vapor boundary (as with $\text{CO}_2$ below its triple point).

For better accuracy, you can measure $P$ at several temperatures, plot $\ln P$ vs. $1/T$, and determine $L$ from the slope of the best-fit line rather than relying on just two data points.

Limitations of the Clausius-Clapeyron Equation

The equation works well for pure substances at moderate conditions, but each approximation introduces limits:

• Constant $L$ assumption. Latent heat actually varies with temperature. For water, $L_{vap}$ drops from about 45 kJ/mol near 0°C to about 40.7 kJ/mol at 100°C. Over narrow temperature ranges this doesn't matter much, but over wide ranges it introduces error.
• Negligible condensed-phase volume. The assumption $v_v \gg v_l$ breaks down at high pressures and especially near the critical point, where the liquid and vapor densities converge and $\Delta v \to 0$.
• Ideal gas behavior. Real vapors deviate from $Pv = RT$ at high pressures or when strong intermolecular forces are present (e.g., hydrogen bonding in water vapor). At those conditions, you'd need a more realistic equation of state.
• Pure substances only. The equation doesn't account for the effect of dissolved solutes on vapor pressure. For solutions, you need modifications like Raoult's law (for ideal solutions) or activity-based corrections.